Question

The self gravitational potential energy of a spherical shell of mass m and radius r is

Answer 1

Answer:

The general formula to get the potential energy of any spherical distribution is this :

U=−∫R0GM(r)rρ(r)4πr2dr,(1)

where M(r) is the mass inside a shell of radius r<R. It is easy to get the gravitational energy of a uniform sphere of mass M and radius R :

U=−3GM25R.(2)

In general, for any spherical distribution of total mass M and exterior radius R, we can write this :

U=−kGM2R,(3)

where k>0 is a constant that depends on the internal distribution. k=35 for the uniform distribution. For a thin spherical shell of radius R (all mass concentrated on its surface), we can get k=12.

Now, I suspect that for all cases :

12≤k<∞.(4)

Physically, this makes sense. But how to prove this from the general integral (1) ?

To simply things a bit, we may introduce the dimensionless variable x=r/R≤1, and defines relative mass M¯(x)≡M(r)/M≤1 and relative density ρ¯(x)=ρ(r)/ρaverage, where ρaverage=3M/4πR3. Thus, integral (1) takes the following form :

U=−3GM2R∫10M¯(x)ρ¯(x)xdx.(5)

The last integral is k3. I'm not sure this may help to prove (4).

Explanation:

Answer 2

Answer:

The self gravitation potential energy of a spherical shell of mass m and radius R is -gm*2/2r

Explanation: